I made a note some time ago that I had read in some book that the equation


has positive integer solutions, but that these are so large that it would be absolutely hopeless to search for them with computers. Unfortunately, I didn't write down where I read this and if you only have the equation, the results Google gives you aren't very helpful. I could only find this so far.

Can someone point me to an article or book where I can read more about this equation? (Preferably something with a proof of the claim above which is accessible even if you're not an expert in number theory.)

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    $\begingroup$ +1.Not relevant to this specific equation, but an interesting (and deeply frustrating :P) fact: such "titanic" Diophantine equations are guaranteed to exist, in a precise sense. For any computable function $f$, there is some $n$ and some Diophantine equation in $k\le n$ variables with all coefficients $\le n$ which has a solution, but no solution in $[0, f(n)]^k$ - that is, we have to look "incomputably" high for solutions to general Diophantine equations. This is a consequence of the solution to Hilbert's Tenth Problem. $\endgroup$ Jan 15, 2016 at 13:42
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    $\begingroup$ Actually, you can find the titanic$x,y$ using computers and some clever shortcuts. A first step is solving $p^3+q^3 = 313^2$ which can then be birationally transformed into the elliptic curve $w^3-432\cdot313^4 = t^2$. If someone would be so kind as to find an initial rational point on this curve, then it is only routine machine manipulation to yield positive integer $x,y$. $\endgroup$ Jan 15, 2016 at 15:29
  • $\begingroup$ Magma says the curve does have a rational point, but since the online version is limited to 120 secs, it does not spell it out. Tease. :( $\endgroup$ Jan 15, 2016 at 15:44
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    $\begingroup$ @TitoPiezasIII According to this page, an integral solution for $p^3 + q^3 = 313^2 r^3$ is $$\begin{align} p: & {\small +355507307842882624593086325021133856149447336710120844428552934573043094018915289363},\\ q: & {\small -354602746692986709129018423204648314355484458881941451025238387384142099383045862152},\\ r: & {\small +1517122651849438712721950935044230084378368307868200665761294465082177989014675811} \end{align} $$ $\endgroup$ Jan 15, 2016 at 16:19
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    $\begingroup$ @TitoPiezasIII amazing, that's probably too big for the margin. $\endgroup$ Jan 15, 2016 at 17:33

2 Answers 2


The equation $313(x^3+y^3) = t^3$ is equivalent to finding,


As Achille Hui points out, Noam Elkies found the solution with $83$-digits,

$$\small x_0 = 355507307842882624593086325021133856149447336710120844428552934573043094018915 289363\\ \small y_0 = -354602746692986709129018423204648314355484458881941451025238387384142099383045 862152 \\ \small z_0 =1517122651849438712721950935044230084378368307868200665761294465082177989014675811$$

Update: Curiously, $3(x_0+y_0) = (3\cdot 7\cdot 8273\cdot 64072783\cdot 125303678787043)^3$.

However, the OP wants them positive. Using the method also discussed in this post, given an initial solution to,

$$ax^3+by^3 = cz^3\tag2$$

then a new one can be derived as,

$$a(-bxy^3-cxz^3)^3 + b(ax^3y+cyz^3)^3 = c(-ax^3z+by^3z)^3\tag3$$

We can use $(3)$ iteratively to find an infinite number of solutions. We have,

$$x_0,y_0,z_0 = +,\,-,\,+\\ x_1,y_1,z_1 = +,\,-,\,-\\ x_2,y_2,z_2 = +,\,-,\,+\\ x_3,y_3,z_3 = +,\,-,\,-\\ x_4,y_4,z_4 = \color{red}{+,\,+,\,+} $$

and so on. So the fourth iteration is all positive. Approximately,

$$x_4 = 1.908757×10^{21389}\\ y_4 = 4.955536×10^{21389}\\ z_4 = 1.095063×10^{21388}$$

They are too long to explicitly write down, but if you have Mathematica, you can retrace the steps taken and see those numbers in all their glory.

  • $\begingroup$ This is really a much surprising result and a beautiful answer $\endgroup$ Jan 15, 2016 at 19:08
  • $\begingroup$ Is it known that this method gives us the minimal positive solution? OP mentions (citing a source) that it is known that this equation has no solutions which are significantly shorter, but I don't see how your approach would establish that. Don't get me wrong - I nevertheless consider your post to be a good one, I'm just wondering if the other result follows from it. $\endgroup$
    – Wojowu
    Jan 17, 2016 at 21:31
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    $\begingroup$ @Wojowu: The source refers to titanic numbers as having 1000 or more decimal digits, so this is our lower bound. However, the method described in this post depends on an initial solution. So while it shows an easy way to generate an infinite number of answers, it may also skip others, hence its first positive solution is not necessarily the smallest. As MacLeod points out, there is in fact a smaller one with about $6770$ digits. $\endgroup$ Jan 19, 2016 at 14:14
  • $\begingroup$ @TitoPiezasIII I see. I thought that titanic numbers are ones with 10000 digits or more. Thanks for your reply. $\endgroup$
    – Wojowu
    Jan 19, 2016 at 14:15
  • $\begingroup$ @Wojowu: I've just asked MacLeod if positive points $mP$ have a predictable $m$. $\endgroup$ Jan 19, 2016 at 14:18

The curve $a^3+b^3=N$ is birationally equivalent to the elliptic curve $y^2=x^3-432N^2$ with \begin{equation} a=\frac{36N+y}{6x} \hspace{2cm} b=\frac{36N-y}{6x} \end{equation}

For $N=313^2$, the curve has rank $1$ and generator $G=(x_0,y_0)$ with



which give the solution Noam Elkies tabulated.

The curve has no torsion points so all rational solutions come from points of the form $mP, \, \, m=1,2,\ldots$.

Looking at the above transformations, a positive solution will only arise when $|y| < 36N$ since $x>0$ always.

Experiments show that this first happens when $m=9$ giving a solution with roughly 6770 digits

  • $\begingroup$ Since the point $mP$ for $m=9$ is positive, are such $m$ predictable? Are these $m$ in arithmetic progression (for example)? $\endgroup$ Jan 19, 2016 at 14:17
  • $\begingroup$ The isogeny $P\to 9P$ has degree $81$ Do you know the story $2^{64}$ attached to chess? Numbers involved are greater than $2$ and with exponent greater than $64$!!! $\endgroup$
    – Piquito
    Jan 20, 2016 at 1:08
  • $\begingroup$ As far as I am aware, one cannot predict where the first positive solution will occur. It all depends on the initial generator and the underlying geometry, $\endgroup$ Jan 22, 2016 at 5:39

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